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2 years ago

Complete the equation and tell which property you used (18x2)x5=18x(2x____)

1 answer:

4 0

<span>Associative Property.
(18x2)x5=18x(2x5)
All you have to do is move the parentheses but keep the numbers the same.</span>

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Leo sold candy bars for a fundraiser. He spent $25 on a box of candy bars and sold them for $2 each. His profit (after earning b

kykrilka [37]

As given by the question

There are given that the $2c-$25=$69.

Now,

$\begin{gathered} $2c-$25=$69$ \\ $2c-$25+25=$69$+25 \\ 2c=$69$+25 \\ 2c=94 \\ c=\frac{94}{2} \\ c=47\text{ bars} \end{gathered}$

Hence, the correct answer is C.

4 0

1 year ago

Yuhhhhhhhhhhhhhhhhh

Mashutka [201]

Answer:

6. 550

7. 70

Step-by-step explanation:

6. 16+28 = 44, 44*25 = 1100 divided by 2 = 550

7. 2.5+3+5.5 = 11, 11+3 = 14, 14*10 = 140 divided by 2 = 70

3 0

2 years ago

A group of 5 men are meeting for lunch.

Sladkaya [172]

Tile:

<h2>See the explanation.</h2>

Step-by-step explanation:

(a)

There were total 5 men wearing coats.

5 coats can be returned to 5 men in 5! = 120 ways.

The coats can be returned to the accurate persons only in 1 way.

Hence, the probability that each man gets the correct coat is $\frac{1}{120}$ .

(b)

At the time of returning the first coat, the hostess will have 5 choices and for the second she will have 4 choices.

Hence, in $5\times4 = 20$ ways the hostess can return the 2 coats.

There is only 1 possible case that each of the coats will return to the correct owner.

Hence, the required probability is $\frac{1}{20}$ .

6 0

2 years ago

Consider a system with one component that is subject to failure, and suppose that we have 115 copies of the component. Suppose f

castortr0y [4]

Answer:

Step-by-step explanation:

From the given information:

the mean $(\mu) = 115 \times 20$

= 2300

Standard deviation = $20 \times \sqrt{115}$

Standard deviation (SD) = 214.4761

TO find:

a) $P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})$

$P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})$

$P(x > 3500)= P(Z > \dfrac{1200}{214.4761})$

$P(x > 3500)= P(Z >5.595)$

From the Z-table, since 5.595 is > 3.999

$P(x > 3500)=1-0.9999$

P(x > 3500) = 0.0001

Here, the replacement time for the mean $(\mu) = \dfrac{0+0.5}{2}$

= 0.25

Replacement time for the Standard deviation $\sigma = \dfrac{0.5-0}{\sqrt{12}}$

$\sigma = 0.1443$

For 115 component, the mean time = (115 × 20)+(114×0.25)

= 2300 + 28.5

= 2328.5

Standard deviation = $\sqrt{(115\times 20^2) +(114\times (0.1443)^2)}$

= $\sqrt{(115\times 400) +(114\times 0.02082249}$

= $\sqrt{(46000) +2.37376386}$

= $\sqrt{(46000) +(2.37376386)}$

= $\sqrt{46002.374}$

= 214.482

Now; the required probability:

$P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})$

$P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})$

$P(x > 4125) = P(Z >8.376)$

$P(x > 4125) =1- P(Z$

From the Z-table, since 8.376 is > 3.999

P(x > 4125) = 1 - 0.9999

P(x > 4125) = 0.0001

7 0

2 years ago

Please help i will mark brainly

Vsevolod [243]

The answer to your question is positive

8 0

2 years ago

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